# How could one integrate ln sin x

## Partial integration

### Repetition: product rule as derivation rule

The **partial integration** is a technique for integrating special functions. These functions consist of the product of two functions, their **Derivatives** are known.

In order to better understand the derivation of the partial integration, you have to refer to the **Product rule** recall. This is one **Derivation rule**. We use it to derive the product of functions $ u (x) \ cdot v (x) $:

$ \ left (u (x) \ times v (x) \ right) '= u' (x) \ times v (x) + u (x) \ times v '(x) $.

In your **Formula collection** you might also find the abbreviation $ (uv) '= u'v + uv' $.

### Derivation of the partial integration

With the help of **Product rule** can the **partial integration** can be derived: The product rule says that two functions are equal: $ \ left (u (x) \ cdot v (x) \ right) '$ is the same as $ u' (x) \ cdot v (x) + u ( x) \ cdot v '(x) $.

So they are also correct **indefinite integrals** these functions match. We see this by integrating both sides after the variable. ("We just put an integral in front of it on both sides.")

$ \ begin {array} {lll} \ int \ left (u (x) \ cdot v (x) \ right) 'dx & = & \ int \ left (u' (x) \ cdot v (x) + u (x) \ cdot v '(x) \ right) dx \ & = & \ int \ left (u' (x) \ cdot v (x) \ right) dx + \ int \ left (u (x) \ cdot v '(x) \ right) dx \ end {array} $

Since the **Differentiation** and the **integration** cancel each other out, i.e. $ \ int f '(x) dx = f (x) $, we can transform it as follows:

$ u (x) \ times v (x) = \ int \ left (u '(x) \ times v (x) \ right) dx + \ int \ left (u (x) \ times v' (x) \ right ) dx $

Now we subtract the term $ \ int \ left (u (x) \ cdot v '(x) \ right) dx $ on both sides of the equation:

$ u (x) \ cdot v (x) - \ int \ left (u (x) \ cdot v '(x) \ right) dx = \ int \ left (u' (x) \ cdot v (x) \ right) dx $

For a better overview, let's swap the two sides of the equation:

$ \ int \ left (u '(x) \ cdot v (x) \ right) dx = u (x) \ cdot v (x) - \ int \ left (u (x) \ cdot v' (x) \ right) dx $

The rule shown here is called **partial integration** or **Product integration**. The following abbreviated form is easier to remember:

$ \ int (u'v) dx = uv- \ int (uv ') dx $

### When do you use partial integration?

The **partial integration** we apply to integrate a function that consists of two or more factors. Examples of such functions are:

- $ f (x) = x \ cdot e ^ x $
- $ g (x) = \ sin (x) \ cdot \ cos (x) $
- $ h (x) = x \ cdot \ ln (x) $

### What should you watch out for with partial integration?

You can see that on both sides of the **partial integration** is an integral. What should be an advantage here? We want it to be easier and not harder!

For this purpose, when integrating a product function, we have to consider which of the two factors “plays the role of $ u '(x) $”. From this function we also need the **Indefinite integral** know. The other function "then plays the role of $ v (x) $".

### example

Let us consider the function $ f (x) = x ^ 2 \ cdot x $. Of course, you can also integrate this function more easily using the power rule of integration. Namely, $ f (x) = x ^ 2 \ cdot x = x ^ 3 $ and thus:

$ \ int f (x) dx = \ frac14x ^ 4 + c $

However, this example is well suited to explain the technique of partial integration and to illustrate its validity. For the partial integration we now assume the following functions:

We also need the derivative of $ v (x) $ and the function $ u (x) $:

- $ u (x) = \ frac 13 x ^ 3 $
- $ v '(x) = 1 $

This results in the following calculation:

$ \ begin {array} {lll} \ int \ underbrace {x ^ 2} _ {u '(x)} \ cdot \ underbrace {x} _ {v (x)} dx & = & \ underbrace {\ frac13x ^ 3} _ {u (x)} \ cdot \ underbrace {x} _ {v (x)} - \ int \ underbrace {\ frac13x ^ 3} _ {u (x)} \ cdot \ underbrace {1} _ { v '(x)} dx \ & = & \ frac13x ^ 4- \ frac1 {12} x ^ 4 + c \ & = & \ frac3 {12} x ^ 4 + c \ & = & \ frac14 x ^ 4 + c \ end {array} $

This antiderivative with $ c \ in \ mathbb {R} $ corresponds to the antiderivative above, which we determined using the power rule.

### Exercises on the application of partial integration

In the following, we will use a few examples to look at which of the two factors of the product function to be integrated should play the role of $ u '(x) $ and how we can use the **partial integration** finally apply.

### Exponential functions

We want to calculate the indefinite integral $ \ int (x \ cdot e ^ x) dx $.

We choose $ u '(x) = e ^ x $ and $ v (x) = x $ because the antiderivative of $ u' (x) $, namely $ u (x) = e ^ x $, is known and because $ v '(x) = 1 $ is easy to use.

At the **Derive** For polynomials, the exponent is always $ 1 $ smaller. That's why you're voting for **Exponential functions** of the above form always the polynomial as $ v (x) $ and the exponential factor as $ u '(x) $. This is often called a **Clearing polynomials** designated.

Now we can **partial integration** apply.

$ \ begin {array} {lll} \ int (x \ cdot e ^ x) dx & = & x \ cdot e ^ x- \ int (1 \ cdot e ^ x) dx \ & = & x \ cdot e ^ xe ^ x + c \ & = & (x-1) \ cdot e ^ x + c \ end {array} $

Undoubtedly the right integral is easier to calculate. If we want to integrate $ f (x) = x ^ 2 \ cdot e ^ x $, we proceed in the same way. However, it must be partially integrated twice.

### Trigonometric functions

A **common area of application** for the partial integration are the trigonometric functions. Let's take as **example** the function $ f (x) = \ sin (x) \ cdot \ cos (x) $. Let's first determine what we will need later:

- $ u '(x) = \ sin (x) $
- $ u (x) = - \ cos (x) $
- $ v (x) = \ cos (x) $
- $ v '(x) = - \ sin (x) $

Then we can start calculating.

$ \ begin {array} {lll} \ int (\ sin (x) \ cdot \ cos (x)) dx & = & - \ cos (x) \ cdot \ cos (x) - \ int ((- \ cos (x)) \ cdot (- \ sin (x))) dx \ & = & - (\ cos (x)) ^ 2- \ int (\ sin (x) \ cdot \ cos (x)) dx \ end {array} $

As we can see, the starting integral appears again. If we add $ \ int (\ sin (x) \ cdot \ cos (x)) dx $ on both sides, we get

$ \ begin {array} {lll} 2 \ int (\ sin (x) \ cdot \ cos (x)) dx & = & - (\ cos (x)) ^ 2 & | &: 2 \ \ int (\ sin (x) \ cdot \ cos (x)) dx & = & - \ frac12 (\ cos (x)) ^ 2 \ end {array} $

### Logarithmic functions

Finally, we use a trick to integrate logarithmic functions. Let's consider the simple example $ f (x) = \ ln (x) $. In order to partially integrate this function, we add the factor $ 1 $, since we need a product: $ \ int \ ln (x) dx = \ int (1 \ cdot \ ln (x)) dx $. Then the following applies:

- $ u '(x) = 1 $
- $ u (x) = x $
- $ v (x) = \ ln (x) $
- $ v '(x) = \ frac 1x $

The partial integration then looks like this:

$ \ begin {array} {lll} \ int \ ln (x) dx & = & \ int (1 \ cdot \ ln (x)) dx \ & = & x \ cdot \ ln (x) - \ int \ left (x \ cdot \ frac1x \ right) dx \ & = & x \ cdot \ ln (x) - \ int 1dx \ & = & x \ cdot \ ln (x) -x + c \ end {array} $

Since the derivative of $ \ ln (x) $ is just $ \ left (\ ln (x) \ right) '= \ frac1x $, $ x $ and $ \ frac1 {x} $ cancel each other out. Therefore you always choose the polynomial as $ u '(x) $ and the logarithm as $ v (x) $.

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